 Hi and welcome to the session. Let us discuss the following question. The question says diagonal AC and BD of quadrilateral ABCD intersect at O in such a way that area of triangle AOD is equal to area of triangle BoC. Prove that ABCD is a trapezius. Let's now make a figure of this question. This is the quadrilateral ABCD diagonals AC and BD intersect at O in such a way that area of triangle AOD is equal to area of triangle BoC. We have to prove that ABCD is a trapezius. Let's now begin with the solution. We are given that area of triangle AOD is equal to area of triangle BoC. Let's name this as equation number one. Now we will add area of triangle BoC on both sides of one. In area of triangle BoC on both sides of one, we get area of triangle AOD plus area of triangle DoC is equal to area of triangle BoC plus area of triangle BoC. Now triangle AOD plus triangle DoC is equal to triangle ACD. So this means area of triangle AOD plus area of triangle DoC is equal to area of triangle ACD. Triangle BoC plus triangle DoC is equal to triangle BCD. So area of triangle BoC plus area of triangle DoC is equal to area of triangle BCD. Let's name this as two. Now since triangle ACD BCD on the same base CD have equal areas triangle BCD live between the same parallel AB and CD. My theorem 9.3 given in your book which states that two triangles in the same base have equal areas live between the same parallel. Here to CD, therefore, ABCD is a trapezium which says that a coordinate rule in which exactly six sides are called trapezium. We have proof that ABCD is a trapezium. This completes the session. Bye and take care.